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Homework Help: Simple Harmonic Motion of an earthquake

  1. Dec 4, 2007 #1
    [SOLVED] Simple Harmonic Motion

    1. The problem statement, all variables and given/known data

    Four people, each with mass of 71.1 kg, are in a car with a mass of 1180 kg. An earthquake strikes. The vertical oscillations of the ground surface make the car bounce up and down on its suspension springs, but the driver manages to pull of the road and stop. When the frequency of the shaking is 1.60 Hz, the car exhibits a maximum amplitude of vibration. The earthquake ends and the four people leave the car as fast as they can. By what distance does the car's undamaged suspension lift the car's body as the people get out?

    2. Relevant equations

    or 2pisqrt(k/m)
    k= 4pi^2m/T^2

    3. The attempt at a solution

    I have notice that mass must be all added to together to give a total mass of 1446.4 kg. Then I found the period by T= 1/f=1/1.60=0.625s or is it T= 2pi/w= 2pi/1.60=3.93s, ^anyway from that i used K=4pi^2m/T^2 to come up with k constant k=4pi^2(1446.4)/(0.625)^2= 147999 n/m , i dont know soemthing tells me im not doing this right i dont how i could go about getting the postion. can someone help me out?
  2. jcsd
  3. Dec 5, 2007 #2

    Shooting Star

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    Homework Helper

    (w means omega.)

    When the driving frequency was f=1.6 Hz, there was resonance, which means w^2= k/m => (2*pi*f)^2 = k/M, where M = mass of car+men.

    delta_F = k*delta_x => Mg - weight of men = delta_x, which is what you want.
  4. Dec 5, 2007 #3
    hmm some calculations i did from this were w^2=k/m (2*pi*1.6)^2=k/1464.4= 147992.264 K= 101.06*1464.4= 147992.264 n/m ok now i tired your equation delta_F=k*delta_x=> Mg-weight of men= delta_x so (1464.4)(9.80)-(71.1*4)= delta x= 14066.72 this was wrong. I also notice what happen to k?
  5. Dec 6, 2007 #4

    Shooting Star

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    What happened to k was a typo: it should be k*delta_x.

    Let's go through the logic once more. The natural frequency of the loaded car must be f =1.6 Hz. That gives us
    (2*pi*f)^2 = k/(M+m), where m is the sum of the masses of the four men.

    Suppose x1 is the compression when the load is M+m, and x2 when load is M. Then,
    (M+m)g – mg = k(x2-x1) => Mg = k*delta_x => delta_x = Mg/[(M+m)(2*pi*f)^2].

    (In my last post, I wrote (Mg - wt of men) by mistake.)

    I get delta_x as 0.078 m.
    Last edited: Dec 6, 2007
  6. Dec 6, 2007 #5
    thanks man that was right. i got same thing :). I appreicate the help.
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